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Questions related to the spectrum of graphs, defined using one of the possible variants of the discrete Laplace operator or Laplacian matrix. See https://en.wikipedia.org/wiki/Discrete_Laplace_operator
2
votes
signing a strongly regular graph
Here are a few ideas on places to look for examples. You may not find (m)any this way.
First to eliminate a trivial case: For some people a complete graph or disjoint union of isomorphic complete gra …
1
vote
Characteristic polynomial of hypercube graph
It is interesting, worthwhile, and not very difficult to understand the situation for the Cartesian product of two or more arbitrary graphs. However here is a simplified answer which applies to this c …
2
votes
Accepted
largest adjacency eigenvalue of line graphs
The lower bound is less than $$\frac{\deg_{\max}}2+1$$ but not by much.
Here is (part of) a graph with an edge adjacent to $10$ others and (part of) the corresponding line graph.
The largest eigen …
1
vote
Spectral gap for random bipartite regular graphs
I assume you mean to fix $d$ and let $n$ grow. When $d \gt 3$ the graph (at least in the case that all vertices have degree $r$) will likely be connected and even $d$-connected. But in case $d=2$ one …
3
votes
Extreme Laplacian eigenvalues
Here is something which seems to work. Consider a circulant graph with vertices $v_0,\cdots,v_{2^n-1}$ and $v_i$ adjacent to $v_{i\pm d}$ where $d$ ranges over a set $D$ of $\lceil\frac{n+1}{2}\rceil …
1
vote
Accepted
How to identify bridge nodes between nearly connected graph components in partitioned adjace...
As i commented, you need to be more specific about the details. Here is an idea though which might work well if within groups there are lots (or at least a reasonable number of) triangles but no trian …
2
votes
Non-isomorphic graphs with the same numbers of closed walks
Look around. The term cospectral is also used. Some of the people who have been answering you showed that Almost all Trees are Co-spectral. Of course trees can be quite irregular. An early paper is Co …
1
vote
Graphs whose degree vectors coincide for all powers of their adjacency matrices
As I mentioned in a comment. Two $n$-vertex graphs , both regular of degree $d$ , will have this property, although the graphs can be quite different in some ways, for example one may be connected and …
2
votes
On the spectrum of random regular graph
I guessed no but Gordon has changed my mind for degree greater than 2.
If it has degree 2 it is a union of cycles. Eigenvalues are $2\cos(\frac{2 \pi j}{k})$ for various $ k$. In particular 2 has mu …
2
votes
equitable partitions
The question is fairly broad. I can say a number of things, but none of them are very deep. For a regular graph the largest eigenvalue of the adjacency matrix, $A$, is the common degree with eigenvect …
1
vote
Questions about interlacing polynomials
It has nothing essential to do with polynomials. Just sets of numbers.
Let $\beta_{11} \lt \beta_{21} \lt \cdots \beta_{n1}$ and $\beta_{12} \lt \beta_{22} \lt \cdots \beta_{n2}$ be two lists. They h …
2
votes
Accepted
Coherence between different ranking methods of a graph's vertices
I'm not sure how common this is. It may depend on how you are determining random graphs.
There may be ties so let me separate the issues of the equivalence relation "same rank" from those of the lin …
5
votes
Accepted
Classes of graphs for which isospectrum implies isomorphism?
Maximum degree 2 would be such a class (which includes regular of degree $2$ as a subclass). Transitive graphs (by which I mean that the relation of being connected by an edge is transitive) are anoth …
3
votes
Generalised Isospectrality of Graphs
Here is a ridiculous solution using a $1 \times 1$ matrix! First, encode an $n \times n$ adjacency matrix $A$ by $a \lt 2^{\binom{n}2}$ where $a$ is the binary integer obtained by listing the above di …
5
votes
How do eigenvalues of combinatorial Laplacian relates to automorphisms in graphs?
For $v=6k+1$ or $v=6k+3$ and $k$ not extremely small there are a great many ways to make a Steiner Triple System with $v$ objects and $M=\frac{v(v-1)}6$ triples so that each pair of objects is in a un …